# Using the fact that <msqrt> n </msqrt> is an irrational number whenever n is not

Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show 3$\sqrt{3}+\sqrt{7}+\sqrt{21}$ is irrational.
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cefflid6y
If $\left(1+\sqrt{3}\right)\left(1+\sqrt{7}\right)$ is rational, then
$\frac{12}{\left(1+\sqrt{3}\right)\left(1+\sqrt{7}\right)}=\frac{12\left(1-\sqrt{3}\right)\left(1-\sqrt{7}\right)}{\left(-2\right)\left(-6\right)}=1-\sqrt{3}-\sqrt{7}+\sqrt{21}$
is also rational.
So, $\frac{1}{2}\left[\left(1+\sqrt{3}\right)\left(1+\sqrt{7}\right)+1-\sqrt{3}-\sqrt{7}+\sqrt{21}\right]-1=\sqrt{21}$ is rational.
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EnvivyEvoxys6
$\begin{array}{rcl}N=\sqrt{3}+\sqrt{7}+\sqrt{21}& ⇒& N-\sqrt{21}=\sqrt{3}+\sqrt{7}\\ & ⇒& {N}^{2}+21-2N\sqrt{21}=10+2\sqrt{21}\\ & ⇒& \sqrt{21}=\frac{{N}^{2}+11}{2+2N}\end{array}$
So $\sqrt{21}$ is rational, which is a contradiction.